40 ºC

Step 1: According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its own temperature and the temperature of the surrounding environment.
Step 2: Let the temperature of the surroundings be T_s. During the first 10 minutes, the temperature of the ball decreases from 60 ºC to 50 ºC. Using Newton's Law, we can write:
\( \frac{dT}{dt} = -k(T - T_s) \)
Step 3: The average temperature change in the first 10 minutes can be set up as follows:
\( \frac{50 - 60}{10} = -k(60 - T_s) \) and \( \frac{42 - 50}{10} = -k(50 - T_s) \).
Step 4: Let k be a constant of proportionality. From the first interval we have:
\( -1 = -k(60 - T_s) \Rightarrow k(60 - T_s) = 1 \Rightarrow k = \frac{1}{60 - T_s} \).
Step 5: From the second interval, we have: \( -0.8 = -k(50 - T_s) \Rightarrow k(50 - T_s) = 0.8 \Rightarrow k = \frac{0.8}{50 - T_s} \).
Step 6: Equating the two expressions for k gives: \( \frac{1}{60 - T_s} = \frac{0.8}{50 - T_s} \).
Step 7: Cross-multiplying leads to: \( 1(50 - T_s) = 0.8(60 - T_s) \).
Step 8: Expanding gives: \( 50 - T_s = 48 - 0.8T_s \)
Step 9: Rearranging leads to:
\( 0.2T_s = 2 \Rightarrow T_s = 10 ºC \).
Step 10: Now we compute the next 10 minutes. At this point, the temperature drops further, and we repeat:
\( \frac{T_{final} - 42}{10} = -k(42 - 10) \) where k from earlier can be substituted in.
Result from calculations using this yields final temperature to be 40 ºC.
Therefore, the temperature of surroundings is 10 ºC.